Geometry – Finding Z Coordinates of the 3rd Point in a 3D Right Triangle

3dcoordinate systemsgeometrytrianglestrigonometry

this is my first post.. I hope this good

I have 1 triangle in space (3D)… and I know all data except the coordinates of 3er point(vertex)… for example this:

triangle data example

then:

Right Triangle 

    x     y      z

v1 = (1.42, 0.239, 0.032)

v2 = (2.97, 1.500, 0.032)

v3 = (2.33, ???, ????)

A = 2

B = 0.99 = 1

C = 2.232

a = 63.664°

b = 26.336°

c= 90°

with all this data I thought that with a simple trigonometric identities I could achieve the goal…BUT NO!

then I check this:

How to find the third coordinate of a right triangle given 2 coordinates and lengths of each side

and this:
Find the third vertex of a triangle in $3D$

3d geometry: triangle 2 points known, find 3rd point

and all information in

How to find coordinates of 3rd vertex of a right angled triangle when everything else is known?

with all that information I found this:
http://awaismunir.net/universal/tangents/3rd-third-vertext-calculate-right-angled-triangle.gif

then I deduced the following:

v3cox = ((-v1.y * opposite) + (v2.y * opposite) + (adjacent * v1.x)) / adjacent

v3coy = ((v1.co * opposite) - (v2.x * opposite) + (adjacent * v1.y)) / adjacent

and this work in 2d!….

then I tried with:

v3coz = ((v1.z * opposite) - (v2.z * opposite) + (adjacent * v1.z)) / adjacent
or
v3coz = ((v1.z * opposite) + (v2.z * opposite) - (adjacent * v1.z)) / adjacent
or
v3coz = ((v1.z * opposite) + (v2.z * opposite) + (adjacent * v1.z)) / adjacent

but any of this "coz" work

I hope you can help me

—————————–
UPDATE

AFTER @Martigan IDEAS I understand that I have infinite number of points that will satisfy the same conditions:

enter image description here

what's up if I know some data about v3.. for example v3x = 2.33 v3y = ?? and v3.z =??

Best Answer

Let $\overrightarrow{B} . \overrightarrow{A}=0$

I call $x_3$ the $x$ coordinate of vertex $v_3$, for convenience purpose.

That is $(x_3-x_2)(x_2-x_1)+(y_3-y_2)(y_2-y_1)+(z_3-z_2)(z_2-z_1)=0$

Given that you know $x_1,x_2,x_3,y_1, y_2, z_1,z_2$ it gives you a relationship between $y_3$ and $z_3$ (or, easier to use - see afterwards, a relationship between $(y_3-y_2)$ and $(z_3-z_2)$.

Then you know that the lenght of $\overrightarrow{B}$ is $0.99$

Then you have $(x_3-x_2)^2+(y_3-y_2)^2+(z_3-z_2)^2=0.99^2$

Granted, this gives you a potential of two values for $y_3$, but you should be able to rule out one of the two if needed.

You could have used the scalar product between $\overrightarrow{B}$ and $\overrightarrow{C}$, that would have given you a non-quadratic equation, but that I feel more complex to compute.

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